You have found the following ages (in years) of all 6 lions at your local zoo: $ 10,\enspace 5,\enspace 8,\enspace 5,\enspace 3,\enspace 16$ What is the average age of the lions at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{10 + 5 + 8 + 5 + 3 + 16}{{6}} = {7.8\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $10$ years $2.2$ years $4.84$ years $^2$ $5$ years $-2.8$ years $7.84$ years $^2$ $8$ years $0.2$ years $0.04$ years $^2$ $5$ years $-2.8$ years $7.84$ years $^2$ $3$ years $-4.8$ years $23.04$ years $^2$ $16$ years $8.2$ years $67.24$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{4.84} + {7.84} + {0.04} + {7.84} + {23.04} + {67.24}} {{6}} $ $ {\sigma^2} = \dfrac{{110.84}}{{6}} = {18.47\text{ years}^2} $ The average lion at the zoo is 7.8 years old. The population variance is 18.47 years $^2$.